A cylinder is closed at both the ends by 30 , | vedclass

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(A) The solid consists of a cylinder and two cones at its ends.Radius of cylinder $(r)$ = $21 \, cm$.Height of each cone $(h_{cone})$ = $30 \, cm$.Tot

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$138600$

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(A) The solid consists of a cylinder and two cones at its ends.Radius of cylinder $(r)$ = $21 \, cm$.Height of each cone $(h_{cone})$ = $30 \, cm$.Total height of the solid = $140 \, cm$.Height of the cylinder $(h_{cyl})$ = Total height - $2 	imes h_{cone} = 140 - 2(30) = 140 - 60 = 80 \, cm$.Volume of the solid = Volume of cylinder + $2 	imes$ Volume of cone.Volume of cylinder = $\pi r^2 h_{cyl} = \frac{22}{7} 	imes 21 	imes 21 	imes 80 = 22 	imes 3 	imes 21 	imes 80 = 1,10,880 \, cm^3$.Volume of two cones = $2 	imes (\frac{1}{3} \pi r^2 h_{cone}) = 2 	imes \frac{1}{3} 	imes \frac{22}{7} 	imes 21 	imes 21 	imes 30 = 2 	imes 22 	imes 21 	imes 30 = 27,720 \, cm^3$.Total Volume = $1,10,880 + 27,720 = 1,38,600 \, cm^3$.

$A$ cylinder is closed at both the ends by $30 \, cm$ high cones. The radius of the cylinder is $21 \, cm$ and the total height of the solid is $140 \, cm$. Find the volume of this solid (in $cm^3$).

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